What Are the Values of k That Satisfy This Differential Equation?

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SUMMARY

The discussion centers on solving the differential equation 4y" = -25y to determine the values of k for which the function y = cos(kt) is a solution. Participants clarify that substituting y = cos(kt) into the equation leads to the identification of two specific values of k, which are k = ±5/2. Additionally, it is emphasized that verifying solutions does not require solving the entire equation but rather substituting values to check their validity.

PREREQUISITES
  • Understanding of differential equations, specifically second-order linear equations.
  • Familiarity with trigonometric functions and their derivatives.
  • Knowledge of the method of substitution in solving equations.
  • Basic skills in calculus, particularly in relation to the work of James Stewart.
NEXT STEPS
  • Study the method of solving second-order linear differential equations.
  • Learn about the characteristics of trigonometric functions in calculus.
  • Explore the concept of verifying solutions through substitution in differential equations.
  • Review James Stewart's "Single Variable Calculus" for additional examples and explanations.
USEFUL FOR

Students of calculus, particularly those struggling with differential equations, as well as educators seeking to provide clearer explanations of solution verification methods.

sristi
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I missed class last Thursday and I am completely confused about the homework. I tried reading the text but there aren't any example problems like this one. Anyone know any free resources for James Stewart Single Variable Calculus?

Well here is the question:

For what values of k does the function y=cos kt satisfy the differential equation 4y"=-25y.
For those values of k, verify that every member of the family of functions y=A sin kt + B cos Kt.

I tried solving for y" and setting the equations equal to each other. To tell the truth I don't understand what the question is saying.

Thanks for any help!
 
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Have you tried to solve the equation directly

y'' = -k^2*cos(kt)
y = cos(kt)

than you get two values of k +/-.
 
For a problem where you are asked "show that A is a solution to problem B", actually solving the problem is overkill. Suppose you were asked to show that x= 1 is a solution to the equation x5- 3x4+ 2x3- 4x2+ 3x- 1= 0. Would you solve the equation or just substitute 1 for x and see if the resulting equation is true?

sristi, substute cos(kt) into 4y"= 5y and see what happens. For what values of k is the resulting equation true?
 

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